Question Number 195027 by York12 last updated on 22/Jul/23
$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}} \\ $$$${find}\:\left({a}_{\mathrm{1}} \right) \\ $$
Commented by mr W last updated on 25/Jul/23
$${i}\:{got}\:{a}_{\mathrm{1}} =\frac{\mathrm{77}}{\mathrm{15}} \\ $$
Commented by York12 last updated on 25/Jul/23
$${please}\:{sir}\:{explain}\:{why} \\ $$
Answered by mr W last updated on 25/Jul/23
$$\Rightarrow\alpha+\beta+\gamma=\frac{\mathrm{1}}{{a}_{\mathrm{3}} } \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\frac{{a}_{\mathrm{1}} }{\mathrm{7}}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{7}}\left(\frac{\mathrm{1}}{{x}}\right)−\frac{{a}_{\mathrm{3}} }{\mathrm{7}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{a}_{\mathrm{1}} }{\mathrm{7}}=\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma} \\ $$$$ \\ $$$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{7}={x}\left({a}_{\mathrm{3}} {x}^{\mathrm{2}} +{a}_{\mathrm{1}} \right) \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{1}}{{k}},\:{say} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha\left({a}_{\mathrm{3}} \alpha^{\mathrm{2}} +{a}_{\mathrm{1}} \right)}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta\left({a}_{\mathrm{3}} \beta^{\mathrm{2}} +{a}_{\mathrm{1}} \right)}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma\left({a}_{\mathrm{3}} \gamma^{\mathrm{2}} +{a}_{\mathrm{1}} \right)}=\frac{\mathrm{1}}{{k}} \\ $$$$\Rightarrow{a}_{\mathrm{3}} \alpha+\frac{{a}_{\mathrm{1}} }{\alpha}=\mathrm{225}{k} \\ $$$$\Rightarrow{a}_{\mathrm{3}} \beta+\frac{{a}_{\mathrm{1}} }{\beta}=\mathrm{144}{k} \\ $$$$\Rightarrow{a}_{\mathrm{3}} \gamma+\frac{{a}_{\mathrm{1}} }{\gamma}=\mathrm{100}{k} \\ $$$$\Rightarrow{a}_{\mathrm{3}} \left(\alpha+\beta+\gamma\right)+{a}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}\right)=\mathrm{469}{k} \\ $$$$\Rightarrow\mathrm{1}+\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{7}}=\mathrm{469}{k} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\sqrt{\mathrm{7}\left(\mathrm{469}{k}−\mathrm{1}\right)} \\ $$$$ \\ $$$$\frac{\mathrm{225}}{\mathrm{1}+\frac{\mathrm{7}}{\alpha^{\mathrm{2}} }}=\frac{\mathrm{144}}{\mathrm{1}+\frac{\mathrm{7}}{\beta^{\mathrm{2}} }}=\frac{\mathrm{100}}{\mathrm{1}+\frac{\mathrm{7}}{\gamma^{\mathrm{2}} }}=\frac{\mathrm{1}}{{k}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{7}}}{\alpha}=\sqrt{\mathrm{225}{k}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{7}}}{\beta}=\sqrt{\mathrm{144}{k}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{7}}}{\gamma}=\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\sqrt{\mathrm{7}}\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}\right)=\sqrt{\mathrm{225}{k}−\mathrm{1}}+\sqrt{\mathrm{144}{k}−\mathrm{1}}+\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\frac{{a}_{\mathrm{1}} }{\:\sqrt{\mathrm{7}}}=\sqrt{\mathrm{225}{k}−\mathrm{1}}+\sqrt{\mathrm{144}{k}−\mathrm{1}}+\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\Rightarrow\sqrt{\mathrm{469}{k}−\mathrm{1}}=\sqrt{\mathrm{225}{k}−\mathrm{1}}+\sqrt{\mathrm{144}{k}−\mathrm{1}}+\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\Rightarrow\sqrt{\mathrm{225}{k}−\mathrm{1}}+\sqrt{\mathrm{144}{k}−\mathrm{1}}=\sqrt{\mathrm{469}{k}−\mathrm{1}}−\sqrt{\mathrm{100}{k}−\mathrm{1}} \\ $$$$\Rightarrow\sqrt{\left(\mathrm{225}{k}−\mathrm{1}\right)\left(\mathrm{144}{k}−\mathrm{1}\right)}=\mathrm{100}{k}−\sqrt{\left(\mathrm{469}{k}−\mathrm{1}\right)\left(\mathrm{100}{k}−\mathrm{1}\right)} \\ $$$$\Rightarrow\mathrm{245}{k}−\mathrm{2}=\mathrm{2}\sqrt{\left(\mathrm{469}{k}−\mathrm{1}\right)\left(\mathrm{100}{k}−\mathrm{1}\right)} \\ $$$$\Rightarrow\mathrm{127575}{k}=\mathrm{1296} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1296}}{\mathrm{127575}}=\frac{\mathrm{16}}{\mathrm{1575}} \\ $$$${a}_{\mathrm{1}} =\sqrt{\mathrm{7}\left(\frac{\mathrm{469}×\mathrm{16}}{\mathrm{1575}}−\mathrm{1}\right)}=\frac{\mathrm{77}}{\mathrm{15}}\:\checkmark \\ $$$$ \\ $$$$\alpha=\sqrt{\frac{\mathrm{7}}{\mathrm{225}×\frac{\mathrm{16}}{\mathrm{1575}}−\mathrm{1}}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\beta=\sqrt{\frac{\mathrm{7}}{\mathrm{144}×\frac{\mathrm{16}}{\mathrm{1575}}−\mathrm{1}}}=\frac{\mathrm{35}}{\mathrm{9}} \\ $$$$\gamma=\sqrt{\frac{\mathrm{7}}{\mathrm{100}×\frac{\mathrm{16}}{\mathrm{1575}}−\mathrm{1}}}=\mathrm{21} \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{1}}{\frac{\mathrm{7}}{\mathrm{3}}+\frac{\mathrm{35}}{\mathrm{9}}+\mathrm{21}}=\frac{\mathrm{9}}{\mathrm{245}} \\ $$
Commented by York12 last updated on 25/Jul/23
$${thanks}\:{so}\:{much} \\ $$